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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2020 · Mathematics · Basics of Mathematics

If x= _c(a b), y= _a(b c), z= _b(c a) , then which of the following...

If x=log⁡c(ab),y=log⁡a(bc),z=log⁡b(ca)x=\log _c(a b), y=\log _a(b c), z=\log _b(c a), then which of the following is correct?

Options

  1. Axyz=1x y z=1
  2. Bx+y+z=1x+y+z=1
  3. C(1+x)−1+(1+y)−1+(1+z)−1=1(1+x)^{-1}+(1+y)^{-1}+(1+z)^{-1}=1
  4. D(1+x)−2+(1+y)−2+(1+z)−2=1(1+x)^{-2}+(1+y)^{-2}+(1+z)^{-2}=1

Correct answer

C. (1+x)−1+(1+y)−1+(1+z)−1=1(1+x)^{-1}+(1+y)^{-1}+(1+z)^{-1}=1

Step-by-step solution

1+x=log⁡c(ab)+1=log⁡c(ab)+log⁡cc=log⁡c(abc)(1+x)−1=1log⁡c(abc)=log⁡(abc)c\begin{aligned} 1+x & =\log _c(a b)+1 \\ & =\log _c(a b)+\log _c c=\log _c(a b c) \\ (1+x)^{-1} & =\frac{1}{\log _c(a b c)}=\log _{(a b c)} c \end{aligned}(1+x)−1=1log⁡c(abc)=log⁡(abc)c(1+x)^{-1}=\frac{1}{\log _c(a b c)}=\log _{(a b c)} c Similarly, (1+y)−1=log⁡(abc)a(1+y)^{-1}=\log _{(a b c)} a and (1+z)−1=log⁡(abc)b(1+z)^{-1}=\log _{(a b c)} b Now, (1+x)−1+(1+y)−1+(1+z)−1(1+x)^{-1}+(1+y)^{-1}+(1+z)^{-1}=log⁡(abc)c+log⁡(abc)a+log⁡(abc)b=log⁡(abc)(cab)=1\begin{aligned} & =\log _{(a b c)} c+\log _{(a b c)} a+\log _{(a b c)} b \\ & =\log _{(a b c)}(c a b)=1\end{aligned}$$
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